Basically the same as Formatting Sandbox in Meta Stack Exchange, but since this and Statistical Analysis are the only two sites (I know) supporting $\TeX$ formatting, I believe we also need one here for testing it.

$\begingroup$I would understand the newly covered elements at the time we choose $S_3$ should be $S_3\setminus(S_1\cup S_2)$ instead of $S_3\setminus(S_1\cap S_2\cap S_3)$.$\endgroup$
– Apass.JackAug 25 '18 at 20:51

43 Answers
43

A suggestion: if you want to see you TeX previewed, pretend to type your question/answer. Then wait for 4 seconds. We have on the fly previewing for LaTeX here. This way we don't keep popping this question to the top of meta.

I am testing whether there are any uses of the #rrggbb color notation to represent usefully distinguishable colors. Certainly $\col{#d10000}$ is distinguishable from $\col{#df0000}$, but the former is indistinguishable from $\col{#d00}$ and the latter from $\col{#e00}$.

Conclusion: on a typical LCD monitor, a half-step (#08) is perceptible in the lighter colors, but not in the darker ones. Even a full step (#11) is too small to be useful for distinguishing different text in a post on this web site.

$\begingroup$Note that whether or not those colors are distinguishable depends upon the capabilities of the monitor and graphics system. Most consumer level displays have limited capabilities (8-bit,low gamut). For some discussion see e.g. here.$\endgroup$
– Bill DubuqueJan 16 '14 at 1:47

$\begingroup$@KennyTM Thanks! Of course it must be so, or else #FFF wouldn't be white. Thanks for pointing this out.$\endgroup$
– MJDJan 16 '14 at 14:13

$\begingroup$@Bill That is interesting, but not relevant to the issue of mathematical typesetting on this web site.$\endgroup$
– MJDJan 16 '14 at 15:32

$\begingroup$@MJD Sure it is. You are attempting to judge if color differences are perceptible on MSE. My point is that it is very difficult to accurately judge that unless one has professional-level graphics hardware and specialized knowledge in this area. For example, what you see as different may display the same to someone else using a monitor with less capability (e.g. one using dithering/interpolation from 8bit to 10bit color).$\endgroup$
– Bill DubuqueJan 16 '14 at 15:48

$\begingroup$I am trying to judge if color differences are usefully different. For example, you are fond of using colored text to highlight parts of equations, as here. The fact that $\color{#0000c3}{\text{#0000c3}}$ might be distinguishable from $\color{#0000cc}{\text{#0000cc}}$ on a professional-quality wide-gamut monitor is of absolutely no use to you in doing that.$\endgroup$
– MJDJan 16 '14 at 15:52

$\begingroup$My goal in writing this post was to decide if I should mention the #rrggbb notation in this post about typesetting colors, in addition to the #rgb notation. My conclusion is that there is no need to do that.$\endgroup$
– MJDJan 16 '14 at 15:55

Now hiding a whole paragraph - again everything has to go in a single line for this to work:

Since $N$ is fixed, we have some amount of primes $$p_1 < p_2 < \ldots p_k \leq N $$ We also have for any $x\in A\setminus\{1\}$ $$x = p_1^{l_1}p_2^{l_2}\ldots p_k^{l_k} ,\quad l_i\geq 0$$ Now it gets rough: By fixing the pair $(m,n)$ we have: $$m=p_1^{s_1}p_2^{s_2}\ldots p_k^{s_k},\ s_i\geq 0\qquad n=p_1^{t_1}p_2^{t_2}\ldots p_k^{t_k}, t_i\geq 0 $$ So we start counting powers $$\left (\begin{array}{}s_1 & s_2 &\ldots & s_k\\t_1 & t_2 &\ldots &t_k \end{array}\right ) $$ For neither $m$ nor $n$ can we have all the primes represented with power $\geq 1$, since that immediately makes the other number equal to $1$, which we have omitted for now.

$\begingroup$AHA! Martin's name did not appear when I tagged him!$\endgroup$
– ahornMar 30 '15 at 9:21

$\begingroup$@ahorn See meta.math.stackexchange.com/questions/6281/… and other related threads. (BTW I think that the correct word in this context is to ping and not to tag.) And if you are wonrdering, the notification from your comment reached me.$\endgroup$
– Martin SleziakMar 30 '15 at 9:36

$\begingroup$Adding to what @Quixotic said in his/her comment, just add \limits in front of \sum, of which can be applied to other similar commands like \prod (which generates $\prod$ lest you did not know).$\endgroup$
– Mr PieApr 8 at 6:44

More are to be added, you can contribute by making any flag you want :-)

$\begingroup$Consider replacing all of your \color{white} with \color{#}. The same can be done with other colors such as replacing \color{darkblue} with \color{#009}.$\endgroup$
– BradJun 30 '14 at 6:14

I asked myself something like "can we know everything in Math?" and found an answer at Quora:

Mathematics is so great it can even answer this question :). Kurt Gödel answered this question almost a century ago with Gödel's incompleteness theorems. ... we will never be able to answer all questions.

Fine, so math is infinite and we will never know everything, but

what kind of infinity is it? Is math or the set of mathematical theorems a ordered infinite set?

If it is a countable infinite set, how to prove that?

I ask if it possible to apply the concept of (infinite) sets onto mathematical theorems. I look at mathematical theorems as elements of a set, and I ask is this set infinite...

EDIT
A way to formalize this could look as follows:

Let $S=\{a_0,a_1,...;t_0,t_1,...\}$ be a set. It contains axioms $a_n$ and theorems $t_m$ that are unprovable with the given axioms. You'll examples as answers to this question.

From a Gödel point of view the set is incomplete, but it could be extended by a, lets call it Gödel operation $\mathfrak G$ that maps
$$
\begin{array}{cl}
\mathfrak G:& S\to S \\
& \{a_0,...,a_k;t_0,...t_m\} \mapsto \{a_0,...,a_k;t_0,...,t_m,t_{m+1}\}
\end{array}
$$
by extending $S$ with an axiom for fixes a hole in the landscape of proofs.
Now if you apply $\mathfrak G$ several times it looks like you can enumerate the individuals elements of $S$, finally making it a countable infinite number of theorems that make up your set.

Test for MathJax trickeries on hyperlinks where the tag appears on the LEFT. Also see the closely related post about footnotes using html.

The main objective is to get the tag (visual cue) on the left, at the beginning of a line, be it in mathmode or text (sectioning).

Does this work? Yes, and inline \ref{sec01} tag works \ref{sec02} just as well in a standard mode or \ref{keypoint}. Now the challenge is to $\hspace{600pt}\ref{sec03}$. Hell yeah! .... umm why is it one line off? See the thoughts on phantom lines later.

So let's see how these silly ones work. First there's the blank Eq.\eqref{05} and the 2nd Eq.\eqref{12}, both tags consisting of 3 white spaces. Then we have Eq.\eqref{13} and lastly Eq.\eqref{14}. oh how about not in eqref but jus ref like these Eq.\ref{05}, Eq.\ref{12}, Eq.\ref{13}, and Eq.\ref{14}?

So this is a hyperlink to Eq.\,\eqref{01} which is supposed to be the normal tag, and then Eq.\,\eqref{02} is supposed to clash with Eq.\,\eqref{03}. Of course, the standard use is like Eq.\,\eqref{04}. oh damn it no thin space unless in mathmode so maybe just Eq.\eqref{04} unless one goes through the trouble of $\text{Eq.}\,\eqref{04}$? Yeah there IS a visible diffrence.

Now this is getting interesting: \Eq.\eqref{21}, or should I call \Eq.\eqref{22}? Getting into the realm of absurdity with \Eq.\eqref{23} and oh btw \Eq.\eqref{24} also with negative that's gonna mess things up, and \Eq.\eqref{25} is no exception. With just \ \ref we have \Eq.\ref{21} ipsum oh I cannot remember \Eq.\ref{22} yes lorem ipsum it is \Eq.\ref{23} can I block call this thing \Eq.\ref{24} by \require \Eq.\ref{25}?

Let's quote without the push \Eq.\eqref{31} and with the deliberate \Eq.$\hspace{100pt}\eqref{31}$ positive space paired with the equation labels \Eq.$\hspace{2000pt}\eqref{32}$, Oh my god I cannot believe this works. But is that the inevitable double parenthesis? How about \Eq.$\hspace{300pt}\eqref{33}$, is there still this thing? Now this \Eq.$\hspace{400pt}\eqref{34}$ and again \Eq.$\hspace{500pt}\eqref{35}$ and on and on to \Eq.$\hspace{600pt}\eqref{36}$. Hold on let's see $\hspace{100pt}\ref{31}$, $\hspace{200pt}\ref{32}$, $\hspace{300pt}\ref{33}$. Yeah, these show that $\hspace{400pt}\ref{34}$ tag position push $\hspace{500pt}\ref{35}$ is not affected by equation content, and we have $\hspace{600pt}\ref{36}$. Okay so there's no need to manually tag (label) the parenthesis just one-sided? Now do these $\hspace{700pt}\ref{37}$ go over the boarder $\hspace{800pt}\ref{38}$? Yes they totally do. So it is capped at 600? How about $\hspace{600pt}\ref{37}$ and $\hspace{600pt}\ref{38}$? Yes, exactly 600pt….. wait, there’s a HUGE problem to this system with any negative space: the page jumps to the middle (or below) the equation and cutting the letters in half, unlike the standard setup that jumps to show the entire equation regardless of the height of the expression. The immediate response to solve this is to always setup a phantom line just above $\tag*{wherever}\label{keypoint}$ you want to jump to (ha, not exactly like this), and maybe use some commands the squash the height of the phantom line. This is less than

$\Large\color{green}{\text{I guess for the sake of sectioning}\ldots} \tag*{Sec.1} \label{sec01}$

. . . it’s okay to setup a phantom line (the whole section heading is in mathmode anyway).

$\begingroup$That's interesting. So $\text{does this $x^2$ also not work properly?}$ Well, it seems to work.$\endgroup$
– celtschkJul 14 '12 at 16:06

1

$\begingroup$MathJax properly handles nested dollars, but they are not protected from MarkDown when used in comments. They are when used in questions and answers.$\endgroup$
– Davide CervoneJul 14 '12 at 18:30

$\begingroup$Do you mean that in one dimension the definition of a manifold with boundary is: "A topological space such that each point has a neighbourhood homeomorphic to an open subset of $(-\infty, 0]$ or $[0, \infty)"? Or do you mean you can define a one dimensional manifold with boundary as a topological space such that each point has a neighbourhood homeomorphic to an open subset of $(-\infty, 0]$, or equivalently, $[0, \infty)$? Surely you mean the latter as $\phi : [0, \infty) \to (-\infty, 0]$, $\phi(x) = -x$ is a homeomorphism.$\endgroup$
– Martin SleziakOct 12 '12 at 13:36

Now without any changes they are put together and everything is broken:
$\|e^A\|=\|I+A+A^2/2!+A^3/3!+\ldots\|=\sup_{|x|=1}\|(I+A+A^2/2!+A^3/3!+\ldots)x\|= \sup_{|x|}\|Ix+Ax+(A^2x)/2!+(A^3x)/3!+\ldots\|\leq \sup_{|x|}\|Ix\|+\sup_{|x|=1}\|Ax\|+\frac{\sup_{|x|}\|A^2x\|}{2!}+\frac{\sup_{|x|}\|A^3x\|}{3!}+\ldots$

This first part of the fromula still works ok:
$\|e^A\|=\|I+A+A^2/2!+A^3/3!+\ldots\|=\sup_{|x|=1}\|(I+A+A^2/2!+A^3/3!+\ldots)x\|= \sup_{|x|}\|Ix+Ax+(A^2x)/2!+(A^3x)/3!+\ldots\|\leq$

When I add the next bit, it becomes weird:
$\|e^A\|=\|I+A+A^2/2!+A^3/3!+\ldots\|=\sup_{|x|=1}\|(I+A+A^2/2!+A^3/3!+\ldots)x\|= \sup_{|x|}\|Ix+Ax+(A^2x)/2!+(A^3x)/3!+\ldots\|\leq \sup_{|x|}\|Ix\|+\sup_{|x|=1}\|Ax\|+$